Transformers, DC Machines, Induction Machines and Synchronous Machines

Section 5 of the GATE Electrical Engineering (EE) paper, and the one every power engineer eventually spends the most working life with. Start with the single-phase transformer's equivalent circuit and the two tests (open-circuit, short-circuit) that populate it without ever loading the machine to destruction, then three-phase transformer connections and the autotransformer; then DC machines, the oldest and conceptually simplest electromechanical energy converter, and how their speed is controlled; then three-phase and single-phase induction machines, the workhorse of industry; and finally synchronous machines, which alone among rotating machines run at exactly the supply frequency and alone can supply or absorb reactive power on demand.

1. Single-phase and three-phase transformers, autotransformer

A single-phase transformer's equivalent circuit (referred to one side) is a series resistance and leakage reactance for each winding, a shunt magnetising branch (core loss resistance in parallel with magnetising reactance), and an ideal transformer of the turns ratio in between; the phasor diagram shows the small magnetising current lagging the applied voltage by nearly 90 degrees, well ahead of the load current it accompanies. The open-circuit (OC) test, run at rated voltage with the secondary open, gives the core (iron) loss and the magnetising-branch parameters, since no load current flows to cause copper loss; the short-circuit (SC) test, run at reduced voltage with the secondary shorted so rated current flows, gives the copper loss and the series impedance, since the reduced voltage means core loss is negligible.

Voltage regulation is the percentage rise in secondary terminal voltage from full load to no load (at constant primary voltage), and depends on the load's power factor as well as its magnitude — it can even be negative for a leading (capacitive) power factor load, where the secondary voltage at full load exceeds its no-load value. Efficiency is maximum at the load where copper loss (which varies as load-current-squared) equals the constant core loss, not at full load — a transformer sized for a load that runs well below rating is deliberately designed to hit this condition near its typical operating point.

ℹ️ Three-phase transformer connections, vector groups and the autotransformer
A three-phase transformer bank's primary and secondary can each be star or delta, giving Yy, Yd, Dy and Dd connections; a Dy or Yd connection introduces a fixed 30-degree phase shift between primary and secondary line voltages, while Yy and Dd introduce none — this is exactly what a vector group label (like Dyn11) records, and two transformers can only be paralleled if their vector groups are compatible. An autotransformer uses a single winding, tapped, instead of two electrically isolated windings, which saves copper and core material for a given kVA rating but sacrifices the electrical isolation a two-winding transformer provides between primary and secondary.

2. Electromechanical energy conversion and DC machines

Electromechanical energy conversion is governed by the same two equations in every rotating machine: the induced EMF is proportional to flux and speed, and the developed torque is proportional to flux and armature current — everything else (transformers excluded, which convert without motion) is detail layered on these two proportionalities. A DC machine's field can be separately excited (an independent supply), shunt (field in parallel with armature) or series (field in series with armature), and the same machine runs as a motor (electrical in, mechanical out) or a generator (mechanical in, electrical out) depending only on which side supplies energy — the equations do not change.

DC motor characteristics by field connection
TypeSpeed-torque behaviourTypical use
Shuntnearly constant speed, moderate starting torquemachine tools, fans
Seriesspeed falls sharply as load rises, very high starting torquetraction, cranes (never run unloaded)
Separately excitedspeed controlled independently of armature currentprecision speed control drives
⚠️ A series motor must never run unloaded
A series motor's field current equals its armature (load) current, so at very light load the flux collapses and speed — inversely proportional to flux for a given back-EMF — rises without bound, a runaway that can mechanically destroy the machine; this is why a series motor is always coupled to its load through a fixed connection (gearing, direct coupling) rather than a belt that can slip off. Speed control of DC motors uses armature voltage control (below base speed, constant torque available) or field flux control (above base speed, constant power available, since weakening the field raises speed but the torque available falls to keep power roughly constant).

3. Three-phase and single-phase induction machines

A three-phase induction motor's stator winding produces a rotating magnetic field at synchronous speed Ns = 120f/P; the rotor, which never turns at exactly Ns (there would then be no relative motion to induce anything), runs at slip s = (Ns - N)/Ns, typically a few percent at rated load. The equivalent circuit per phase mirrors a transformer's, with the rotor branch's resistance split into R2 (actual rotor resistance) plus R2(1-s)/s (a fictitious resistance representing the mechanical power converted) — this split is exactly how the equivalent circuit accounts for shaft output separately from rotor copper loss.

The no-load test (motor run unloaded at rated voltage) gives the core loss and magnetising-branch parameters, exactly parallel to a transformer's OC test; the blocked-rotor test (rotor held stationary, reduced voltage applied so rated current flows) gives the rotor and stator series impedance, exactly parallel to the SC test — the same logic transfers because the induction motor's equivalent circuit IS a transformer's, with the secondary (rotor) circuit closed on itself through a slip-dependent resistance instead of an external load. Starting methods (star-delta starter, auto-transformer starter, soft starter) all attack the same problem: a directly started induction motor draws 5 to 7 times rated current at standstill, since slip = 1 there gives the smallest possible rotor-branch resistance.

🎯 Why a single-phase induction motor cannot self-start
A single-phase winding alone produces a pulsating (not rotating) magnetic field, which can be resolved into two equal and opposite rotating fields — these produce equal and opposite torques at standstill, so the net starting torque is exactly zero. Once the rotor is spun in either direction by an external means, the two component torques become unequal and a net torque sustains rotation in that direction — which is why every practical single-phase induction motor carries a starting mechanism (a capacitor-start winding, a shaded pole) whose only job is to break this symmetry at standstill.

4. Synchronous machines, losses and efficiency

A synchronous machine's rotor runs at exactly synchronous speed Ns = 120*f/P and can be cylindrical (round) rotor, suited to high-speed steam and gas turbine generators with uniform air gap, or salient (projecting) pole, suited to slower hydro generators, whose non-uniform air gap gives it a second, reluctance-based torque component (absent in the cylindrical machine) even with zero field excitation. Voltage regulation of a synchronous generator, defined the same way as a transformer's, is found from the synchronous impedance method (or more refined methods) using the open-circuit and short-circuit characteristics, and rises for a lagging (inductive) load and can go negative for a leading (capacitive) one — the same qualitative behaviour as a transformer, for the same underlying reason.

Two synchronous generators can only be paralleled once four conditions match: same voltage magnitude, same frequency, same phase sequence, and same phase angle at the instant of connection — checked classically with synchronising lamps or a synchroscope. A synchronous motor is not self-starting either (the rotor must already be near synchronous speed before the field is applied, or the reluctance torque and induction-motor-like starting windings called damper/amortisseur windings bring it up first), but once running it holds exactly synchronous speed regardless of load, up to its maximum (pull-out) torque.

🧠 Losses and maximum efficiency, the same rule everywhere
Every electrical machine's losses split into constant losses (core/iron loss, friction and windage — roughly independent of load) and variable losses (copper/I^2R loss — rising with load squared), and maximum efficiency for ANY of them — transformer, DC machine or AC machine alike — occurs at the load where these two loss categories are EQUAL, never at full load and never where total loss is merely 'small'. This single rule, applied consistently, answers most 'at what load is efficiency maximum' questions across the whole of Section 5 without needing a separate derivation for each machine type.

Key takeaways

  • OC test gives core loss and magnetising-branch parameters; SC test gives copper loss and series impedance — for a transformer AND, by the same equivalent-circuit logic, for an induction motor's no-load and blocked-rotor tests.
  • Maximum efficiency of any machine — transformer, DC or AC — occurs where constant (core/friction) loss equals variable (copper) loss, never simply at full load.
  • A series DC motor's flux collapses at light load and its speed rises without bound — never run one unloaded or belt-coupled.
  • A single-phase induction motor's pulsating field resolves into two opposite rotating fields with zero net starting torque, which is exactly why it needs a dedicated starting mechanism and cannot self-start.
  • Two synchronous generators parallel only when voltage, frequency, phase sequence and phase angle all match; a salient-pole machine alone has a reluctance torque component even at zero excitation.

Practice questions (10)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The open-circuit test on a transformer is performed primarily to determine

    1. copper loss and series impedance
    2. core loss and magnetising-branch parameters
    3. the turns ratio only
    4. the insulation resistance
    Show answer

    Answer: B — core loss and magnetising-branch parameters

    With the secondary open, no load current flows, so there is no copper loss to measure — the input at rated voltage is entirely core loss plus the small current that establishes the magnetising branch, which is exactly what this test isolates.
  2. A Dy1 and a Yy0 transformer cannot be operated in parallel chiefly because

    1. their kVA ratings must always differ
    2. they introduce different phase shifts between primary and secondary line voltages
    3. one has a higher efficiency than the other
    4. delta windings cannot carry rated current
    Show answer

    Answer: B — they introduce different phase shifts between primary and secondary line voltages

    A Dy connection introduces a fixed phase shift (a multiple of 30 degrees) between primary and secondary line voltages while a Yy connection introduces none, so paralleling the two would circulate a large current driven by the phase mismatch even if voltage magnitudes matched exactly — vector group compatibility is a precondition for paralleling, independent of rating or efficiency.
  3. A shunt DC motor's field current is suddenly interrupted while it is running unloaded. The motor's speed will

    1. fall to zero immediately
    2. rise dangerously, potentially to a destructive speed
    3. stay exactly the same
    4. reverse direction
    Show answer

    Answer: B — rise dangerously, potentially to a destructive speed

    Speed is inversely proportional to flux for a roughly constant back-EMF, so losing field flux (going toward zero) drives speed toward a dangerously high value — this is the DC-machine analogue of the series-motor no-load runaway, and is why a shunt or separately excited motor's field circuit must never be allowed to open while the armature is energised.
  4. Which of the following are correctly true of DC motor speed control?

    1. Armature voltage control is used below base speed, at constant available torque.
    2. Field flux control is used above base speed, at roughly constant available power.
    3. Weakening the field always increases available torque at a given armature current.
    4. A series motor's speed control cannot use field weakening at all.
    Show answer

    Answer: A — Armature voltage control is used below base speed, at constant available torque.; B — Field flux control is used above base speed, at roughly constant available power.

    The first two are the standard two-region DC drive rule. Weakening the field REDUCES available torque at a given armature current (torque is proportional to flux times current), which is the trade-off for the speed gain, not a free increase; and series motors do use field weakening (a diverter resistance across the series field) as one of several speed-control techniques, so it is not excluded.
  5. A 4-pole, 50 Hz three-phase induction motor's synchronous speed is

    1. 1500 rpm
    2. 3000 rpm
    3. 1000 rpm
    4. 750 rpm
    Show answer

    Answer: A — 1500 rpm

    Ns = 120f/P = 12050/4 = 1500 rpm — a formula that must be applied with P as the number of POLES, not pole pairs, which is the usual place this calculation goes wrong.
  6. In an induction motor's per-phase equivalent circuit, the rotor branch resistance is split as R2 + R2(1-s)/s. The term R2(1-s)/s represents

    1. rotor copper loss
    2. stator core loss
    3. the fictitious resistance representing developed mechanical power
    4. magnetising branch reactance
    Show answer

    Answer: C — the fictitious resistance representing developed mechanical power

    R2 alone is the actual rotor winding resistance, responsible for rotor copper loss; the added term R2(1-s)/s carries no physical resistance of its own but is inserted purely to make the power dissipated in it, in the equivalent circuit, equal the real mechanical power developed at the shaft — separating gross mechanical power from rotor copper loss cleanly.
  7. At the instant a three-phase induction motor is switched on directly (across the line), slip s equals 1, and the current drawn is typically

    1. much lower than rated current
    2. 5 to 7 times rated current
    3. exactly equal to rated current
    4. zero
    Show answer

    Answer: B — 5 to 7 times rated current

    s = 1 makes the fictitious mechanical-power resistance R2(1-s)/s exactly zero, leaving only the small actual rotor resistance R2 in that branch — the smallest possible impedance the rotor branch can present, which is exactly why standstill (direct-on-line) starting current is so much higher than running current, motivating reduced-voltage starters.
  8. A single-phase induction motor at standstill produces net starting torque equal to

    1. the full rated torque
    2. zero, because two equal and opposite rotating field torques cancel
    3. twice the rated torque
    4. a torque in a random, unpredictable direction
    Show answer

    Answer: B — zero, because two equal and opposite rotating field torques cancel

    The pulsating single-phase field resolves into two equal, oppositely rotating fields, each producing an equal-magnitude torque in its own direction at standstill — the two exactly cancel, giving zero net starting torque, which is why an auxiliary starting winding or other means is always needed.
  9. Two three-phase synchronous generators are to be paralleled. Which condition is NOT required before closing the connecting switch?

    1. equal voltage magnitude
    2. equal frequency
    3. equal power output at the instant of connection
    4. same phase sequence and matching phase angle
    Show answer

    Answer: C — equal power output at the instant of connection

    Voltage magnitude, frequency, phase sequence and phase angle must all match before paralleling; the two generators' POWER OUTPUT need not be equal at the instant of connection — that is set afterward by adjusting each machine's prime-mover input, and is not a precondition for closing the breaker.
  10. A transformer's core (constant) loss is 100 W. At the load where its efficiency is maximum, its copper (variable) loss, in watts, is ___

    Numerical answer — type the value.

    Show answer

    Answer: 100

    Maximum efficiency occurs exactly where constant (core) loss equals variable (copper) loss, so at that load the copper loss also equals 100 W — the same rule applies to transformers, DC machines and AC machines alike.